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## #51 2007-04-27 19:25:57

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

#14

Prove that if n is an odd positive integer, there exists a sequence of n consecutive integers whose sum is equal to n[sup]2[/sup].
­

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## #52 2007-04-28 02:47:02

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

#13 I surrender....

Numbers are the essence of the Universe

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## #53 2007-06-01 02:14:54

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

#15 This is an easy one.

Prove that any even power of any odd integer leaves a remainder of 1 when divided by 4.

Last edited by JaneFairfax (2007-06-01 02:15:24)

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## #54 2007-06-01 07:39:14

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

This one is easier

Last edited by Stanley_Marsh (2007-06-01 07:39:28)

Numbers are the essence of the Universe

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## #55 2007-06-01 08:24:51

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

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## #56 2007-06-01 20:01:07

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

Yah , it comes from mutiplication law of mod

Numbers are the essence of the Universe

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## #57 2007-06-23 02:18:53

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

#16

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## #58 2007-08-04 01:22:16

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

#17 Here is another exercise I made up myself.

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## #59 2007-08-07 09:54:03

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

#18

Last edited by JaneFairfax (2007-08-07 10:48:00)

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## #60 2007-08-10 22:14:03

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

This result is important because I used it in proving Tonys question: http://www.mathsisfun.com/forum/viewtopic.php?id=7807

Last edited by JaneFairfax (2007-08-11 03:10:04)

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## #61 2007-08-16 20:38:09

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

#19

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## #62 2007-09-13 11:24:41

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

#20

I just made this problem up myself; hopefully Ive got the maths correct.

Last edited by JaneFairfax (2007-09-13 11:26:02)

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## #63 2007-10-26 22:28:52

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

#21

Consider the sequence

2, 5, 9, 12, 16, 19

The first term is 2 and successive terms are formed by alternately adding 3 and 4 to previous terms:

Prove that

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## #64 2008-02-22 03:44:32

Kurre
Member
Registered: 2006-07-18
Posts: 280

### Re: Janes exercises

a more powerful result of 15:

which is two consecutive even integers, thus one can be written on the form 4k, and one on the form 4k+2, and thus k^{2n}-1 is divisible by 8, ie k^{2n} gives rest 1 when divided by 8.

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## #65 2009-04-14 00:08:05

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 106,389

### Re: Janes exercises

Hi Jane;

Don't suppose anyone will ever see this but here goes:

The recurrence formula for your set of numbers is.

this has a characteristic equation of

x^3-x^2-x+1=0 which has roots of {-1,1,1}

This then is the general form of the solution:

we need to determine c1,c2, and c3 from the initial conditions

So the general solution is:

This agrees with your general solution at the bottom of problem 21

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.

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## #66 2009-04-14 01:15:29

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

bobbym. looks like youve got the answer. Well done!

My solution:

Let

Then

and
are APs with common difference 7 and first terms 2 and 5 respectively. Thus

Note that

can be written as:

The result follows.

Last edited by JaneFairfax (2009-04-14 01:22:51)

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## #67 2009-04-14 01:30:59

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 106,389

### Re: Janes exercises

Hi Jane;

Thanks for showing me yours, I am not that happy with mine, what justification do I have for the first step? I haven't proven that a(n+3)=a(n+2)+a(n+1)-a(n) is the recursion for that sequence. Can you provide some rigor to my arguments?

bobbym

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.

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## #68 2009-04-14 04:36:48

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Well, Im not that familiar with techniques on solving difference equations, but your solution looked right to me so I gave you full credit for it.

Im pretty sure your first step was correct.

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## #69 2009-04-14 11:33:25

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 106,389

### Re: Janes exercises

Hi Jane;

You say you are not  very familiar with methods of solving difference equations but yet you came up with the idea of splitting the sequence into 2 coupled difference equations, each handling every other term. I salute you.

bobbym

Last edited by bobbym (2009-04-14 11:48:04)

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.

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## #70 2009-04-14 12:21:54

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Aw, thanks. Well, I thought up that problem myself, so for this particular problem I happen to have have my own solution handy.

At least, I discovered the following handy lemma myself:

If a sequence A has properties that alternate between its odd and even terms, let B be the subsequence of its odd-numbered terms and C the subsequence of its even-numbered terms. Find formulas for B and C separately. Then combine the formulas by rewriting A in the manner described above.

Id love to call this Janes lemma if no-one else has discovered it before.

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## #71 2009-04-15 01:13:54

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 106,389

### Re: Janes exercises

Hi Jane;

I discovered that about 10 years ago but didn't notice your method of combining them and promptly forgot about the whole idea. So I guess its yours and I will always remember it as Jane's lemma.

bobbym

Last edited by bobbym (2009-04-19 18:24:25)

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.

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